A) A circle of radius = 0.2 m and time period \[\pi \times {{10}^{-7}}s\]
B) A circle of radius = 0.1 m and time period \[2\pi \times {{10}^{-7}}s\]
C) A helix of radius = 0.1 m and time period \[2\pi \times {{10}^{-7}}s\]
D) A helix of radius = 0.2 m and time period \[4\pi \times {{10}^{-7}}s\]
Correct Answer: C
Solution :
Path of the proton will be a helix of radius \[r=\frac{mv\sin \theta }{qB}\] (where q = Angle between \[\overrightarrow{B}\,\text{and}\,\overrightarrow{v\,}\]) \[\Rightarrow \ r=\frac{1.67\times {{10}^{-27}}\times 2\times {{10}^{6}}\times \sin {{30}^{o}}}{1.6\times {{10}^{-19}}\times 0.104}\] \[=0.1m\] Time period \[T=\frac{2\pi m}{qB}=\frac{2\pi \times 1.67\times {{10}^{-27}}}{1.6\times {{10}^{-19}}\times 0.104}\] \[=2\pi \times {{10}^{-7}}sec\]You need to login to perform this action.
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